Residues and Contour Integration Problems

exists and is not 0. We can use L’ H^opital’s rule: lim z!0 zcot(z) = lim z!0 zcos(z) sin(z) = lim z!0 cos(z) zsin(z) cos(z) = 1: Thus the singularity is a simple pole. 2. f(z) = 1+cos(z) (z ˇ)2 at z= ˇ. Ans. Removable. Solution. Power series is the simplest way to do this. We can expand cos(z) in a Taylor series about z= ˇ. To do so ...

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